Question

Let $a_1, a_2, \ldots, a_n$ be in AP If $a_5=2 a_7$ and $a_{11}=18$, then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to

Hint:

In problems involving arithmetic progressions and sums of series, make sure to use the general term formula and simplify terms systematically to calculate the total sum efficiently.

Answer 1

Correct Answer: 8

To solve this problem, let the common difference of the arithmetic progression (AP) be d, and the first term be a. Then, the n-th term of the AP is given by:
a_n = a + (n-1)d.

Given that a₅ = 2a₇ and a₁₁ = 18, we can write:
a + 4d = 2(a + 6d),
a + 10d = 18.

From the first equation, a + 4d = 2a + 12d, simplifying gives:
a = -8d. (Equation 1)

From the second equation:
a + 10d = 18.
Substitute a from Equation 1:
-8d + 10d = 18
2d = 18
d = 9.

Substitute back to find a:
a = -8(9) = -72.

Now, calculate the n-th terms needed:
a₁₀ = a + 9d = -72 + 81 = 9,
a₁₁ = 18,
a₁₂ = a + 11d = -72 + 99 = 27,
a₁₃ = a + 12d = -72 + 108 = 36,
a₁₄ = a + 13d = -72 + 117 = 45,
a₁₅ = a + 14d = -72 + 126 = 54,
a₁₆ = a + 15d = -72 + 135 = 63,
a₁₇ = a + 16d = -72 + 144 = 72,
a₁₈ = a + 17d = -72 + 153 = 81.

The expression to evaluate is:
12(1/(\sqrt{9} + \sqrt{18}) + 1/(\sqrt{18} + \sqrt{27}) + ... + 1/(\sqrt{72} + \sqrt{81})).

Recognizing the pattern, the general term can be simplified by multiplying and dividing by the conjugate:
For consecutive terms a_i and a_i+1:
(\sqrt{a_i+1} - \sqrt{a_i}) / (a_i+1 - a_i).

a_i+1 - a_i = d = 9.

The series simplifies to telescoping terms:
(\sqrt{81} - \sqrt{9})/9 = (\sqrt{81} - \sqrt{9})/9.

- Compute: \(\sqrt{81} - \sqrt{9} = 9 - 3 = 6\).

Final result after multiplying by 12:
12 * (6/9) = 12 * (2/3) = 8.

This computed value is 8, which is the expected and confirmed value within the range.